LOJ166 拉格朗日插值 2【卷积，NTT】

题目链接：LOJ

题目描述：输入多项式的次数$n$，一个整数$m$和$f(0),f(1),f(2),ldots,f(n)$，输出$f(m),f(m+1),f(m+2),ldots,f(m+n)$

数据范围：$1leq nleq 10^5,n<mleq 10^8$

一道披着拉格朗日插值模板的外衣的数论题。。。

$$f(m+i)=sum_{j=0}^nf(j)dfrac{prod_{k ot=j}(m+i-k)}{prod_{k ot=j}(j-k)}$$

$$=dfrac{(m+i)!}{(m-n+i-1)!}sum_{j=0}^ndfrac{f(j)*(-1)^{n-j}}{j!*(n-j)!}*frac{1}{m+i-j}$$

预处理阶乘，逆元，阶乘逆元，和$mfac[i]=prod_{j=0}^{i-1}(m-n+j)$，$m-n+i$的逆元和$mfac[i]$的逆元。

使用NTT计算，时间复杂度为$O(nlog n)$

#include<bits/stdc++.h>
#define Rint register int
using namespace std;
typedef long long LL;
const int N = 1 << 19, mod = 998244353, G = 3, Gi = 332748118;
inline int kasumi(int a, int b){
    int res = 1;
    while(b){
        if(b & 1) res = (LL) res * a % mod;
        a = (LL) a * a % mod;
        b >>= 1;
    }
    return res;
}
int rev[N];
inline int calrev(int len){
    int L = -1, limit = 1;
    while(limit <= len){limit <<= 1; L ++;}
    for(Rint i = 0;i < limit;i ++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << L);
    return limit;
}
inline void NTT(int *A, int limit, int type){
    for(Rint i = 0;i < limit;i ++) if(i < rev[i]) swap(A[i], A[rev[i]]);
    for(Rint mid = 1;mid < limit;mid <<= 1){
        int Wn = kasumi(type == 1 ? G : Gi, (mod - 1) / (mid << 1));
        for(Rint j = 0;j < limit;j += mid << 1){
            int w = 1;
            for(Rint k = 0;k < mid;k ++, w = (LL) w * Wn % mod){
                int x = A[j + k], y = (LL) w * A[j + k + mid] % mod;
                A[j + k] = (x + y) % mod;
                A[j + k + mid] = (x - y + mod) % mod;
            }
        }
    }
    if(type == -1){
        int inv = kasumi(limit, mod - 2);
        for(Rint i = 0;i < limit;i ++) A[i] = (LL) A[i] * inv % mod;
    }
}
int fac[N], invfac[N], inv[N], mfac[N], minvfac[N], minv[N];
inline void init(int n, int m){
    fac[0] = mfac[0] = 1;
    for(Rint i = 1;i <= (n << 1 | 1);i ++){
        fac[i] = (LL) fac[i - 1] * i % mod;
        mfac[i] = (LL) mfac[i - 1] * (m - n + i - 1) % mod;
    }
    invfac[n << 1 | 1] = kasumi(fac[n << 1 | 1], mod - 2);
    minvfac[n << 1 | 1] = kasumi(mfac[n << 1 | 1], mod - 2);
    for(Rint i = (n << 1 | 1);i;i --){
        invfac[i - 1] = (LL) invfac[i] * i % mod;
        minvfac[i - 1] = (LL) minvfac[i] * (m - n + i - 1) % mod;
        inv[i] = (LL) invfac[i] * fac[i - 1] % mod;
        minv[i] = (LL) minvfac[i] * mfac[i - 1] % mod;
    }
    minv[0] = 1;
}
int n, m, f[N], A[N], B[N];
int main(){
    scanf("%d%d", &n, &m);
    for(Rint i = 0;i <= n;i ++) scanf("%d", f + i);
    init(n, m);
    int limit = calrev(n * 3);
    for(Rint i = 0;i <= n;i ++){
        A[i] = (LL) f[i] * invfac[i] % mod * invfac[n - i] % mod;
        if(n - i & 1) A[i] = mod - A[i];
    }
    for(Rint i = 0;i <= (n << 1);i ++) B[i] = minv[i + 1];
    NTT(A, limit, 1); NTT(B, limit, 1);
    for(Rint i = 0;i < limit;i ++) A[i] = (LL) A[i] * B[i] % mod;
    NTT(A, limit, -1);
    for(Rint i = n;i <= (n << 1);i ++)
        printf("%d ", (LL) mfac[i + 1] * minvfac[i - n] % mod * A[i] % mod);
}